Answer:
Step-by-step explanation:
To find x₁ and x₂ :
![\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] \times \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%261%5C%5C5%264%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%5C%5Cx_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step 1
Multiply first 2 x 2 matrix with 2 x 1 vector, we get
![\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2B%20%20x_2%5C%5C5x_1%26%2B%20%204x_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%5C%5C-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 2
Add the 2 x 1 matrices on LHS, we get
![\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4x_1%26%2Bx_2%26%2B11%5C%5C5x_1%26%2B4x_2%26-19%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%5C%5C40%5Cend%7Barray%7D%5Cright%5D)
Step 3,
we get
and
Step 4,
Simplify, we get
Step 5,
multiply eqn(1) by 4
we get
Step 6,
eqn (2) - eqn(3)
we get
substituting in eqn (1), we get
so, we get
Therefore
Answer:
B. A quadrilateral with 2 pairs of equal length
Step-by-step explanation:
analyzing the options:
- A. A polygon is a very general definition (polygon is a figure with 3 or more sides) so it does not specifically describe a rhombus.
- B. This option is much more specific to a rhombus than the other two options because it mentions that it is a quadrilateral so it reduces the generality of calling it a polygon, and it also mentions that there are 2 pairs of equal sides which is true for the rhombus (the top and bottom sides are equal, and the sides are also equal). So this option is correct.
- C. It is a little more specific than option A because it mentions that you must have 4 sides, but it is not as specific for the rhombus as option B
Answer:
188π in^2
Step-by-step explanation:
Surface area of the rocket = surface area of the cone + surface area of the cylinder
Surface area of a cone = πr(r+h^2+r^2)
r = 8/2 = 4 in
4π( 4 + 3^2 + 4^2) =
4π ( 4 + 9 + 16) = 116π in^2
Surface area of a cylinder = 2πrh+2πr^2
(2 x n x 4 x 5) + (2 x n x 16)= 72π in^2
116 + 72 = 188π in^2
Answer:
-7 < -5
Step-by-step explanation:
-5 is greater than -7
